You drive on Interstate 10 from San Antonio to Houston, half the time at $\mathbf{55}\mathbf{km}\mathbf{/}\mathbf{h}$and other half at$\mathbf{90}\mathbf{km}\mathbf{/}\mathbf{h}$. What is your average speed (a) from San Antonio to Houston? (b) from Houston back to San Antonio? (c) for the entire trip? (d) What is your average velocity for the entire trip? (e) Sketch x vs tfor (a), assuming the motion is all in positive x direction. Indicate how the average velocity can be found on the sketch.

The speed in the first half of the trip, ${\mathrm{v}}_1=55\mathrm{km}/\mathrm{h}.$

The speed in the second half of the trip,${\mathrm{v}}_2=90\mathrm{km}/\mathrm{h}$

The average speed from San Antonio can be found using distance relation. The average speed is equal to the total distance traveled divided by the time taken to travel the total distance.

Using average speeds calculated in part (a) and part (b), the average speed for the entire trip can be computed using the average speed equation. And thus, the average velocity for the entire trip can be found using the average velocity equation. The graph of distance vs time can be drawn using speeds given in part (a) for each half time

The average speed can be written as,

$\mathrm{Average}\mathrm{speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}$ (i)

If T is the total time of travel, total distance traveled, D, can be written as,

$\mathrm{D}={\mathrm{V}}_1\left(\frac{\mathrm{T}}{2}\right)+{\mathrm{V}}_2\left(\frac{\mathrm{T}}{2}\right)$

To find the average speed from San Antonio to Houston we can write,

$$\begin{gathered} {\mathrm{S}}_{\mathrm{avg}1}=\frac{\mathrm{D}}{\mathrm{T}} \\ =\frac{55\left({\displaystyle \frac{\mathrm{T}}{2}}\right)+90\left({\displaystyle \frac{\mathrm{T}}{2}}\right)}{\mathrm{T}} \\ =72.5\mathrm{km}/\mathrm{h} \\ \approx 73\mathrm{km}/\mathrm{h} \end{gathered}$$

The average speed is 73km/h.

We can write,

$$\begin{gathered} {\mathrm{S}}_{\mathrm{avg}2}=\frac{\mathrm{D}}{\mathrm{T}} \\ =\frac{\mathrm{D}}{{\displaystyle \frac{\mathrm{D}/2}{55}}+{\displaystyle \frac{\mathrm{D}/2}{90}}} \\ =68.3\mathrm{km}/\mathrm{h} \\ \approx 68\mathrm{km}/\mathrm{h} \end{gathered}$$

Therefore, the average speed from Houston back to San Antonio is 68 km/h

$$\begin{gathered} {\mathrm{S}}_{\mathrm{avg}}=\frac{73+68}{2}\mathrm{The}\mathrm{average}\mathrm{speed}\mathrm{for}\mathrm{entire}\mathrm{trip}\mathrm{is}, \\ =70.4\mathrm{km}/\mathrm{h} \\ \mathrm{The}\mathrm{average}\mathrm{speed}\mathrm{for}\mathrm{the}\mathrm{entire}\mathrm{trip}\mathrm{is}70\mathrm{km}/\mathrm{h}. \end{gathered}$$

For the entire trip net displacement is 0. Therefore, the average velocity for the entire trip is 0 km/hr.

For half time you travel with $55\mathrm{km}/\mathrm{h}$and next half time with$90\mathrm{km}/\mathrm{h}$, covers distance D The average velocity can be found by plotting the slope of the graph which is shown in the figure below.

First-line segment is$\left(0,0\right)$to$\left(\mathrm{T}/2,55\mathrm{T}/2\right)$and the second is$\left(\mathrm{T}/2,55\mathrm{T}/2\right)$to$\left(\mathrm{T},\mathrm{D}\right)$.

Therefore, the average speed from San Antonio to Houston is 73km/h and from Houston to San Antonio is 68km/h.

The average speed for entire trip is 70km/h while average velocity for entire trip is 0km/h.